Question: The increasing sequence of positive integers $a_1,$ $a_2,$ $a_3,$ $\dots$ has the property that
\[a_{n + 2} = a_{n + 1} + a_n\]for all $n \ge 1.$  If $a_7 = 120,$ then find $a_8.$
Solution: Let $a_1 = a$ and $a_2 = b.$  Then
\begin{align*}
a_3 &= a + b, \\
a_4 &= a + 2b, \\
a_5 &= 2a + 3b, \\
a_6 &= 3a + 5b, \\
a_7 &= 5a + 8b, \\
a_8 &= 8a + 13b.
\end{align*}Hence, $5a + 8b = 120.$  Then $5a = 120 - 8b = 8(15 - b).$  Since 5 is relatively prime to 8, $a$ is divisible by 8.

If $a = 8,$ then $b = 10.$  If $a = 16,$ then $b = 5,$ which does not work, because the sequence is increasing, so $b > a.$  Note that higher values of $b$ return lower values of $a,$ so the only possible value of $a$ is 8.  Then $b = 10,$ so $a_8 = 8a + 13b = \boxed{194}.$